Discriminant (Δ) & Nature of Roots — Grade 11 Maths (CAPS)

The discriminant is Δ = b² − 4ac, and its sign tells you the nature of the roots before you solve — two real, one equal, or non-real. Each question type worked in full, with a worksheet.

For a quadratic ax² + bx + c = 0, the part of the quadratic formula under the square root, b² − 4ac, is the discriminant (symbol Δ). It tells you what kind of roots the equation has — without doing the full working. This guide covers the discriminant question types of Grade 11.

The discriminant symbol is Δ, the Greek capital letter delta, so you write Δ = b² − 4ac. Some textbooks use the letter D instead; it means exactly the same thing. Δ is just a number: work it out, look at its sign, and you know what kind of roots the equation has.

1What the discriminant tells you

Δ > 0 : two different real roots (the graph crosses the x-axis twice).
Δ = 0 : one real root, repeated (the graph just touches the x-axis).
Δ < 0 : no real roots (the graph never reaches the x-axis).
When a, b and c are rational — as they are in every exam question — Δ > 0 and a perfect square (1, 4, 9, 16…) : roots are rational and it factorises.
With a, b and c rational, Δ > 0 but NOT a perfect square : roots are real but irrational (surds).

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Why the condition matters: the roots are x = (−b ± √Δ)/2a, so a perfect-square Δ only guarantees rational roots when a and b are rational too. With an irrational coefficient it fails: x² − 2√2x + 1 = 0 has Δ = 8 − 4 = 4, a perfect square, yet its roots √2 ± 1 are irrational.

The discriminant decides how the parabola meets the x-axis: two roots (Δ > 0), one repeated root (Δ = 0), or none (Δ < 0).

2Type 1: Classify the roots (no real roots)

Worked ExampleExample 1: describe the roots of 2x² − 3x + 5 = 0

1
Identify a = 2, b = −3, c = 5.
Match to ax² + bx + c = 0.
2
Δ = b² − 4ac = (−3)² − 4(2)(5) = 9 − 40 = −31.
Mind the sign: (−3)² = +9.
3
Δ = −31 < 0 → no real roots.
A negative under the root has no real value.

3Type 2: Rational vs irrational roots

Worked ExampleExample 2: are the roots of x² − 5x + 6 = 0 rational?

1
Δ = (−5)² − 4(1)(6) = 25 − 24 = 1.
Compute the discriminant.
2
Δ = 1 > 0 and 1 is a perfect square → two rational roots.
Perfect-square Δ means it factorises.

Worked ExampleExample 3: classify the roots of x² − 3x + 1 = 0

1
Δ = (−3)² − 4(1)(1) = 9 − 4 = 5.
2
Δ = 5 > 0 but not a perfect square → two real, irrational roots.
The answer will contain √5.

4Type 3: Equal roots: find k

The common exam twist: an equation contains an unknown (often k) and you must find the value that gives a particular kind of root. Translate the words into a condition on Δ.

Worked ExampleExample 4: for which k does x² + kx + 9 = 0 have equal roots?

1
Equal roots means Δ = 0.
One repeated root ⇒ discriminant is zero.
2
Δ = k² − 4(1)(9) = k² − 36 = 0.
Substitute a = 1, b = k, c = 9.
3
k² = 36 → k = ±6.
Two values give equal roots.

5Type 4: Real roots / non-real roots: find k

Worked ExampleExample 5: for which k does x² − 4x + k = 0 have two real roots?

1
Two real roots means Δ > 0.
Real and distinct ⇒ discriminant positive.
2
Δ = (−4)² − 4(1)(k) = 16 − 4k > 0.
Set up the inequality.
3
16 > 4k → k < 4.
Solve the inequality (divide by 4).

6What equal roots mean

Equal roots means the equation has two roots that are the same number. When Δ = 0 the ± in the quadratic formula adds and subtracts zero, so both roots are x = −b ÷ 2a. On the graph, the parabola touches the x-axis at its turning point instead of crossing it.

Worked ExampleExample 6: show that x² − 10x + 25 = 0 has equal roots

1
Δ = (−10)² − 4(1)(25) = 100 − 100 = 0.
Substitute a = 1, b = −10, c = 25.
2
Δ = 0 → the roots are real and equal.
State the conclusion in words: that is where the mark is.
3
Check: x² − 10x + 25 = (x − 5)² = 0, so x = 5 twice.
The turning point (5; 0) lies on the x-axis.

7Real roots and non-real roots

Real roots are roots you can place on the number line. They exist when Δ ≥ 0, which includes the equal-roots case. Non-real roots happen when Δ < 0: the formula then asks for the square root of a negative number, which is not a real number, and the parabola neither cuts nor touches the x-axis.

Worked ExampleExample 7: show that 3x² + 2x + 1 = 0 has non-real roots

1
Δ = 2² − 4(3)(1) = 4 − 12 = −8.
2
Δ = −8 < 0 → the roots are non-real.
The formula would need √(−8), which is not a real number.

8Type 5: Show the roots are real for every value of k

Worked ExampleExample 8: show that x² − (k + 2)x + 2k = 0 has real roots for every real value of k

1
a = 1, b = −(k + 2), c = 2k.
Keep b in brackets: the whole bracket is squared.
2
Δ = (k + 2)² − 4(1)(2k) = k² + 4k + 4 − 8k = k² − 4k + 4.
3
Δ = (k − 2)² ≥ 0 for every real k.
A square is never negative.
4
Δ ≥ 0 → the roots are always real (and equal when k = 2).
This is the standard way to answer 'for all values of k': write Δ as a perfect square.

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Translate the words into a condition first: 'equal/one repeated root' → Δ = 0; 'two real roots' → Δ > 0; 'real roots' (without 'two') → Δ ≥ 0; 'no real roots' → Δ < 0; 'rational roots' → Δ a perfect square. Then substitute and solve.

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Sign of b²: if b = −3 then b² = 9, not −9. Squaring always gives a positive — this single slip flips many answers.

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Standard form first: in 3x² = 5x − 2 the value of b is not 5. Rewrite it as 3x² − 5x + 2 = 0, so a = 3, b = −5, c = 2 and Δ = 25 − 24 = 1.

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'Real roots' is not 'two real roots': real roots allows equal roots, so the condition is Δ ≥ 0; two unequal real roots needs Δ > 0. In Example 5, 'real roots' would give k ≤ 4, not k < 4.

Because the conditions are fixed and the algebra is short, discriminant questions are among the best-value marks in Grade 11–12 Paper 1. Drill every type below.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Δ for x² − 6x + 9 = 0, and the nature of roots. (classify)
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36 − 36 = 0 → one real (equal) root
2
Δ for x² + 2x + 5 = 0, and the nature of roots. (classify)
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4 − 20 = −16 < 0 → no real roots
3
Δ for 2x² − 7x + 3 = 0, and the nature. (classify)
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49 − 24 = 25 > 0, and 25 is a perfect square → the roots are real, rational and unequal
4
Are the roots of x² − 4x + 4 = 0 rational? (rational?)
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Δ = 0 (perfect square) → rational (equal)
5
Rational or irrational: x² − 2x − 2 = 0? (rational?)
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Δ = 4 + 8 = 12, not a perfect square → irrational
6
For which k does x² + kx + 4 = 0 have equal roots? (equal, find k)
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k² − 16 = 0 → k = ±4
7
For which k does x² − 6x + k = 0 have equal roots? (equal, find k)
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36 − 4k = 0 → k = 9
8
For which k does x² + 2x + k = 0 have two real roots? (real, find k)
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4 − 4k > 0 → k < 1
9
For which k does x² − 2x + k = 0 have no real roots? (non-real, find k)
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4 − 4k < 0 → k > 1
10
Δ for 3x² − 5x + 1 = 0. (compute)
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25 − 12 = 13 (two real, irrational)